Question

How to check the converges of this ..
(n^3-2n+1/n) / (7n^5 +2sin(n)-1 )
Please help me.

Answer 1

convergence as \[n \rightarrow \infty\]?

Answer 2

Hey , Rlsadiz, I need a method to do that, Help me

Answer 3

use limit evaluation as it approaches infinity\[\lim_{n \rightarrow \infty} = \frac{\frac{n^{3} - 2n + 1}{n}}{7n^{5} + 2\sin(n) - 1}\] is this the correct question?

Answer 4

rlsadiz, no the question u typed is wrong.the upper part is n^3-2n+(1/n)

Answer 5

Take out the highest power of n from the numerator and denominator to get (n^3(1-2/n^2+1/n^4))/(n^5(7+2sin(n)/n^5-1/n^5)) You can then ignore anything with an n on the denominator as it approaches zero as n approaches infinity. This leaves you with n^3/7n^5 which equals 1/7n^2, which clearly converges. Hope this is right.

Answer 6

Did you use any tests, eg, Ratio Test or Limit comparison test ..?

Answer 7

That test is for series convergence.

Answer 8

rlsadiz, how do you prove this series is converges then?

Answer 9

What dr25161 did is right.

Answer 10

wait are you asking for \[\sum_{x = 1}^{\infty} the function\]?

Answer 11

rlsadiz, you are right..

Answer 12

I dont know what that test is called ( i learnt calculus in a foreign language) but its the one where you do A(n+1)/A(n) and if its less than 1, the series converges.

Answer 13

dr25161, it is called Ratio test, But I'm stuck in the middle way, Please can you solve it step by step.

Answer 14

You could use comparison test. We need to find a function such that it is larger than our function then prove that it will converge. Consider the function:
\[1.)\frac{n^{3}}{7n^{5} - 3}\] it is larger than our function because
\[-1\le \sin n \le 1\] and \[0\ge-2n+\frac{1}{n}\]
We could now use limit comparison test to evaluate no. 1\[\lim_{n \rightarrow \infty} \left|\frac{\frac{n^{3}}{n^{5} - 3}}{\frac{n^{3}}{n^{5}}} \right|\]
Evaluating the limit by getting the highest power we get 1.
Therefore it converges.

Answer 15

Thanks a lot man, it was really helpful...